Valeur attendue de la médiane de l'échantillon compte tenu de la moyenne de l'échantillon

16

Soit Y la médiane et soit X¯ la moyenne d'un échantillon aléatoire de taille n=2k+1 d'une distribution N(μ,σ2) . Comment puis-je calculer E(Y|X¯=x¯) ?

Intuitivement, en raison de l'hypothèse de normalité, il est logique de prétendre que E(Y|X¯=x¯)=x¯ et c'est effectivement la bonne réponse. Cela peut-il être montré avec rigueur?

Ma pensée initiale était d'aborder ce problème en utilisant la distribution normale conditionnelle qui est généralement un résultat connu. Le problème est que, comme je ne connais pas la valeur attendue et par conséquent la variance de la médiane, je devrais calculer ceux qui utilisent la statistique k+1 er ordre. Mais c'est très compliqué et je préfère ne pas y aller à moins d'y être absolument obligé.

JohnK
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Je pense que c'est une conséquence immédiate de la généralisation que je viens de publier sur stats.stackexchange.com/a/83887 . La distribution des résidus xix¯ est clairement symétrique autour de 0 , d'où leur médiane a une distribution symétrique, donc sa moyenne est nulle. Par conséquent, l'espérance de la médiane elle-même (et pas seulement des résidus) est égale à 0+E(X¯ | X¯=x¯)=x¯ , QED.
whuber
@whuber Désolé, résidus?
JohnK
Je les ai définis dans mon commentaire: ce sont les différences entre chaque xi et leur moyenne.
whuber
@whuber Non, je comprends, mais je travaille toujours à comprendre comment votre autre réponse se rapporte à ma question et comment fonctionne exactement l'attente que vous avez utilisée.
JohnK
2
@whuber D'accord, alors corrigez-moi. Si je me trompe, E(Y|X¯)=E(X¯|X¯)+E(YX¯|X¯) Et maintenant le deuxième terme est zéro parce que le la médiane est symétrique autour de x¯ . Par conséquent, l'attente se réduit à x¯
JohnK

Réponses:

7

Laissez désignent l'échantillon original et Z le vecteur aléatoire avec des entrées Z k = X k - ˉ X . Alors Z est centré normal (mais ses entrées ne sont pas indépendantes, comme le montre le fait que leur somme est nulle avec une probabilité totale). En tant que fonction linéaire de X , le vecteur ( Z , ˉ X ) est normal donc le calcul de sa matrice de covariance suffit à montrer que Z est indépendant de ˉXZZk=XkX¯ZX(Z,X¯)Z .X¯

En ce qui concerne , on voit que Y = ˉ X + TT est la médiane des Z . En particulier, T ne dépend que de Z donc T est indépendant de ˉ X , et la distribution deYY=X¯+TTZTZTX¯Z is symmetric hence T is centered.

Enfin,

E(YX¯)=X¯+E(TX¯)=X¯+E(T)=X¯.
Did
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Thank you, this was asked almost a year ago and I am very glad that someone finally cleared it up.
JohnK
7

The sample median is an order statistic and has a non-normal distribution, so the joint finite-sample distribution of sample median and sample mean (which has a normal distribution) would not be bivariate normal. Resorting to approximations, asymptotically the following holds (see my answer here):

n[(X¯nYn)(μv)]LN[(00),Σ]

with

Σ=(σ2E(|Xv|)[2f(v)]1E(|Xv|)[2f(v)]1[2f(v)]2)

where X¯n is the sample mean and μ the population mean, Yn is the sample median and v the population median, f() is the probability density of the random variables involved and σ2 is the variance.

So approximately for large samples, their joint distribution is bivariate normal, so we have that

E(YnX¯n=x¯)=v+ρσvσX¯(x¯μ)

where ρ is the correlation coefficient.

Manipulating the asymptotic distribution to become the approximate large-sample joint distribution of sample mean and sample median (and not of the standardized quantities), we have

ρ=1nE(|Xv|)[2f(v)]11nσ[2f(v)]1=E(|Xv|)σ

So

E(YnX¯n=x¯)=v+E(|Xv|)σ[2f(v)]1σ(x¯μ)

We have that 2f(v)=2/σ2π due to the symmetry of the normal density so we arrive at

E(YnX¯n=x¯)=v+π2E(|Xμσ|)(x¯μ)

where we have used v=μ. Now the standardized variable is a standard normal, so its absolute value is a half-normal distribution with expected value equal to 2/π (since the underlying variance is unity). So

E(YnX¯n=x¯)=v+π22π(x¯μ)=v+x¯μ=x¯
Alecos Papadopoulos
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As always, nice answer +1. However, since we have no information about the sample size, the asymptotic distribution might not hold. If there is no way to obtain the exact distribution though, I suppose I'll have to make do. Thank you very much.
JohnK
6

The answer is x¯.

Let x=(x1,x2,,xn) have a multivariate distribution F for which all the marginals are symmetric about a common value μ. (It does not matter whether they are independent or even are identically distributed.) Define x¯ to be the arithmetic mean of the xi, x¯=(x1+x2++xn)/n and write xx¯=(x1x¯,x2x¯,,xnx¯) for the vector of residuals. The symmetry assumption on F implies the distribution of xx¯ is symmetric about 0; that is, when ERn is any event,

PrF(xx¯E)=PrF(xx¯E).

Applying the generalized result at /stats//a/83887 shows that the median of xx¯ has a symmetric distribution about 0. Assuming its expectation exists (which is certainly the case when the marginal distributions of the xi are Normal), that expectation has to be 0 (because the symmetry implies it equals its own negative).

Now since subtracting the same value x¯ from each of a set of values does not change their order, Y (the median of the xi) equals x¯ plus the median of xx¯. Consequently its expectation conditional on x¯ equals the expectation of xx¯ conditional on x¯, plus E(x¯ | x¯). The latter obviously is x¯ whereas the former is 0 because the unconditional expectation is 0. Their sum is x¯, QED.

whuber
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Thank you for posting it as a full answer. I now understand the essence of your argument but I might ping you if something is still unclear.
JohnK
5
JohnK, I need to alert you to be cautious. A counterexample to this argument has been brought to my attention. I have encouraged its originator to post it here for further discussion, but briefly it concerns a discrete bivariate distribution with symmetric marginals but asymmetric conditional marginals. Its existence points to a flawed deduction early in my argument. I currently hope that the argument might be rescued by imposing stronger conditions on the xi, but my attention is presently focused elsewhere and I might not get to think about this for awhile.
whuber
4
In the meantime I would encourage you to unaccept this answer. I would ordinarily delete any answer of mine known to be incorrect, but (as you might be able to tell) I like solutions based on first principles rather than detailed calculations, so I hope this argument can be rescued. I therefore intend to leave it open for criticism and improvement (and therefore made it CW); let the votes fall as they may.
whuber
Of course, thanks for letting me know. We will discuss it further when you have time. In the meantime I will settle for the asymptotic argument proposed by @Alecos Papadopoulos.
JohnK
6

This is simpler than the above answers make it. The sample mean is a complete and sufficient statistic (when the variance is known, but our results do not depend on the variance, hence will be valid also in the situation when the variance is unknown). Then the Rao-Blackwell together with the Lehmann-Scheffe theorems (see wikipedia ...) will imply that the conditional expectation of the median, given the arithmetic mean, is the unique minimum variance unbiased estimator of the expectation μ. But we know that is the arithmetic mean, hence the result follows.

We did also use that the median is an unbiased estimator, which follows from symmetry.

kjetil b halvorsen
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By symmetry E[Y]=μ, indeed. Then from these two theorems we know that E[Y|X¯] is the Unique Minimum Variance Unbiased Estimator for μ which we already know to be equal to X¯. This is a brilliant answer, thank you very much. I would have marked it as the correct one, had I not done that already for another answer.
JohnK